Exponent of a finite group $G$

Question

Let $G$ be a finite group. The exponent of $G$, $\exp(G)$ is defined as the minimal positive integer $m$ such that $x^m=1$ for all $x \in G$. prove:

$a)$ if $G$ is abelian then $\exp(G)= \max\{\text{ord}(x):x \in G\}$

$b)$ if $G$ is not abelian the previous statement may fail.

Yes, although it's a bit odd to write $p_i$ to be numbers which are very likely composite... — aleph_two, Dec 20 '18 at 05:41

score 1 · Answer 1 · answered Nov 18 '18 at 02:10

Here is my attempt for

$a)$ $G$ is finite abelian, so $G$ is direct sums of $k$ $\mathbb{Z}/p_i \mathbb{Z}$'s for some $k \in \mathbb{N}$ and $p_i|p_{i+1}$ for all $i$. Hence element with highest order is $(1,...,1)$ with order $lcm(p_1,...,p_k)=p_k$ which is the highest order of any element in $G$ which is $exp(G)$.

$b)$ $S_3$ is a counterexample.

Exponent of a finite group $G$

1 Answers1